Bài 1:

\(a.n_{NaOH\left(tổng\right)}=0,05.1+0,2.0,2=0,09\left(mol\right)\\ V_{ddNaOH\left(tổng\right)}=50+200=250\left(ml\right)=0,25\left(l\right)\\ C_{MddNaOH\left(cuối\right)}=\dfrac{0,09}{0,25}=0,36\left(M\right)\\ b.n_{HCl}=0,5.0,02=0,01\left(mol\right)\\ n_{H_2SO_4}=0,08.0,2=0,016\left(mol\right)\\ V_{ddsau}=20+80=100\left(ml\right)=0,1\left(l\right)\\ C_{MddH_2SO_4}=\dfrac{0,016}{0,1}=0,16\left(M\right)\\ C_{MddHCl}=\dfrac{0,01}{0,1}=0,1\left(M\right)\)

Bài 2:

\(a.m_{H_2SO_4}=29,4.10\%=2,94\left(g\right)\\ b.n_{H_2SO_4}=\dfrac{2,94}{98}=0,03\left(mol\right)\\ n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right)\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Vì:\dfrac{0,01}{1}< \dfrac{0,03}{1}\Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(dư\right)}=0,03-0,01=0,02\left(mol\right)\\ m_{H_2SO_4\left(dư\right)}=0,02.98=1,96\left(g\right)\\ n_{H_2}=n_{Fe}=0,01\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,01.22,4=0,224\left(l\right)\)

\(H_2SO_4+Na_2CO_3\rightarrow Na_2SO_4+CO_2+H_2O\)

\(n_{H_2SO_4}=n_{Na_2CO_3}=0,024.0,1=0,0024\left(mol\right)\)

=> \(CM_{H_2SO_4\left(bđ\right)}=\dfrac{0,0024}{0,002}=1,2M\)

n fe =0.02(mol)  

pthh: fe + h2so4 = feso4 + h2

(mol)0.02 0.02     ->0.02    0.02

=> m feso4 = 0.02 x 152= 3.04 (g)          m dd feso4 = (m fe + m dd h2so4) - m h2

                                                                                    = (1.12 + 200) -(2x 0.02)=201.08(g)

c% feso4 = 1.51%

 bạn nào giúp mình đi chớ giải cả bài mình không hiểu thế vào C% sau PƯ đâu

\(n_{SO_3}=\dfrac{200}{80}=2.5\left(mol\right)\)

\(m_{dd_{H_2SO_4}}=1000\cdot1.12=1120\left(g\right)\)

\(m_{H_2SO_4}=1120\cdot17\%=190.4\left(g\right)\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

\(2.5.....................2.5\)

\(m_{H_2SO_4}=2.5\cdot98+190.4=435.4\left(g\right)\)

\(m_{dd_{H_2SO_4}}=200+1120=1320\left(g\right)\)

\(C\%_{H_2SO_4}=\dfrac{435.4}{1320}\cdot100\%=33\%\)

$SO_3 + H_2O \to H_2SO_4$
$n_{H_2SO_4} = n_{SO_3} = \dfrac{200}{80} = 2,5(mol)$
$m_{dd\ H_2SO_4\ 17\%} =1000.1,12 = 1120(gam)$

Sau khi pha : 

$m_{dd} = 200 + 1120 = 1320(gam)$
$m_{H_2SO_4} = 1120.17\% + 2,5.98 = 435,4(gam)$
$C\%_{H_2SO_4} = \dfrac{435,4}{1320}.100\% = 33\%$

\(n_{SO2}=\dfrac{40}{64}=0,625\left(mol\right)\)

Pt : \(SO_2+H_2O\rightarrow H_2SO_3|\)

         1         1               1

      0,625                    0,625

\(n_{H2SO3}=\dfrac{0,625.1}{1}=0,625\left(mol\right)\)

250ml = 0,25l

\(C_{M_{H2SO3}}=\dfrac{0,625}{0,25}=2,5\left(M\right)\)

 Chúc bạn học tốt

a/ PTHH :      CuO + H2SO4 ===> CuSO4 + H2O

                      0,04         0,04                  0,04                 ( mol )

b/ m_H2SO4= 150 x 10% = 15 gam

=> n_H2SO4= 15 : 98 = 0,15 mol

n_CuO = 3,2 : 80 = 0,04 mol

Theo pt ta thấy CuO pứ hết, H2SO4 dư.

Ta lập tỉ lệ số mol theo pt:

=> m_{CuO pứ}= 3,2 gam

      m_CuSO4= 0,04 x 160 = 6,4 gam

c/ m_{dung dịch thu đc} = 3,2 + 150 = 153,2 gam

n_{H2SO4 dư}= 0,15 - 0,04 = 0,11 mol

=> m_{H2SO4 dư} = 0,11 x 98 = 10,78 gam

=> C%_H2SO4= 10,78 / 153,2 x 100% = 7,03%

      C%_CuSO4= 6,4 / 153,2 x 100% = 4,18%

*Tk

nCuO=3,2/80=0,04(mol)

PTHH: CuO + H2SO4 -> CuSO4 + H2O

nH2SO4= nCuSO4=nCuO=0,04(mol)

=> mH2SO4=0,04. 98=3,92(g)

=> mddH2SO4= 3,92 : 20%=19,6(g)

mddCuSO4= mCuO + mddH2SO4= 3,2+ 19,6= 22,8(g)

mCuSO4=0,04. 160= 6,4(g)

=> C%ddCuSO4= (6,4/22,8).100=28,07%

À mình làm sai, bạn tham khảo bạn Linh nhé!

nguồn :hoc24

nCuO=3,2/80=0,04(mol)

PTHH: CuO + H2SO4 -> CuSO4 + H2O

nH2SO4= nCuSO4=nCuO=0,04(mol)

=> mH2SO4=0,04. 98=3,92(g)

=> mddH2SO4= 3,92 : 20%=19,6(g)

mddCuSO4= mCuO + mddH2SO4= 3,2+ 19,6= 22,8(g)

mCuSO4=0,04. 160= 6,4(g)

=> C%ddCuSO4= (6,4/22,8).100=28,07%